The correct relationship between the `pH` of isomolar solutions of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` and sodium telluride `(pH_(4))` is

Options

A. `pH_4 gt pH_2=pH_3=pH_4`
B. `pH_3 le pH_2 le pH_3 le PH_3 `
C. `pH_4 ge pH_2 le pH_2 `
D. `pH_1 gt pH_2 gt pH_3 gt pH_4`

Answer

D

Explanation

`{:(I , Na_(2)O + H_(2)O to 2 NaOH) , (II , Na_(2)S + H_(2)O to 2NaOH + H_(2)S), (III , Na_(2)Se + H_(2)O to 2 NaOH + H_(2)Se),(IV , Na_(2) Te + H_(2)O to 2NaOH + H_(2)Te):}`
I will have the highest pH , Among `H_(2)S , H_(2) Se , H_(2)` Te acidity goes on increasing on going down the group as bond length increases on increasing size of central atom . So , `pH_(4) lt pH_(3) lt pH_(2)`
Overall order is `pH_(1) gt pH_(2) gt pH_(3) gt pH_(4)`